# $\sin (z+w)= \sin z\cos w +\cos z \sin w \forall z,w \in \mathbb C$ using Identity Theorem

In my Complex Analysis course the following problem was given after teaching "Zeros of analytic function are isolated" and the Identity Theorem. So I was supposed to solve the problem using above theorems.But I could not manage to solve it using these theorems.Please help me to solve the problem. Thnx in advance.

Prove that $\sin (z+w)= \sin z\cos w +\cos z \sin w \forall z,w \in \mathbb C$

What I done so far:

Fixing $w\in \mathbb R$ and considering $f(z)=\sin(z+w),g(z)=\sin z\cos w +\cos z \sin w$ I get that $f$ and $g$ agree on $\mathbb R$ hence they agree on $\mathbb C$. And this is true $\forall w\in \mathbb R$. But I can not figure out how to solve when $w\in \mathbb C$?

I know the problem can be solved just putting values of $\sin,\cos$ in terms of exponential function but I want to solve using the theorems I have learned.

May be I am missing something very easy,I appologise for that.

• I'm not really sure what the problem is.. The identity theorem says exactly that since the two functions agree on a set with an accumulation point, they must be the same. It doesn't matter what the underlying set really is with regards to $\Bbb C$ (only that it is a subset with an accumulation point). – Cameron Williams Sep 13 '14 at 13:28
• Oh I didn't notice that you had $w$ in the problem statement. That was my bad. I guess your concern is that it is going to lead to a cyclic argument. I'll think about it and make an answer shortly. – Cameron Williams Sep 13 '14 at 13:39
• Thank you very much Ian. I am just telling the proof again.Please correct me if I am wrong.So using identity theorem it is true that $\sin (w+x)=\sin w \cos x +\cos w \sin x \forall x \in \mathbb R \forall w\in \mathbb C$ Then I fix $w \in \mathbb C$ to form $f_w(z),g_w(z)$ (as you suugested) and this two agree on $\mathbb R\Rightarrow$ they agree on $\mathbb C$ , And since this is true forall $w\in \mathbb C$ the result follows.Am I right? Thnx again Ian. – usermath Sep 13 '14 at 17:25